#start of lecture 4
## Linear coefficients equations
$$(a_{1}x+b_{1}y+c_{1})dx+(a_{2}x+b_{2}y+c_{2})dy=0 \qquad a_{1},b_{1},c_{1},a_{2},b_{2},c_{2}\in \mathbb{R}$$
> I'm calling this #de_LC_type1 

imagine $c_{1},c_{2}=0$ It becomes a homogenous equation! #de_h_type2

so can we make them 0?
let $x=u+k$
$y=v+l$
where $k,l$ are constants hand picked such that the following terms equal 0:
$(a_{1}u+b_{1}v+\underbrace{\cancel{  c_{1}+a_{1}k+b_{1}l } }_{ 0 })du+(a_{2}u+b_{2}v+\underbrace{ \cancel{ c_{2}+a_{2}k+b_{2}l }  }_{ 0 })dv=0$
In order for these two terms to equal zero, we have to solve this linear system of equations:
$a_{1}k+b_{1}l=-c_1$
$a_{2}k+b_{2}l=-c_{2}$

if $\det\begin{pmatrix}a_{1} & b_{1} \\a_{2} & b_{2}\end{pmatrix}\ne0$ the system is solvable and the DE turns into a homogenous equation.
if $\det\begin{pmatrix}a_{1} & b_{1} \\a_{2} & b_{2}\end{pmatrix}=0 \Rightarrow$ the system is unsolvable but we get an equation of type $\frac{ dy }{ dx }=G(ax+by)$ (also homogenous)

### Example
#ex #de_LC_type1
$$(-3x+y+6)dx+(x+y+2)dy=0$$
let $x=u+k$
$y=v+l$
differentiating we get: $dx=du ,\quad dy=dv$
$(-3u+v+6-3k+l)du+(u+v+2+k+l)dv=0$
we want $6-3k+l$ and $2+k+l$ to equal 0
so:
$-3k+l=-6$
$k+l=-2$
$det\begin{pmatrix}-3 & 1 \\1 & 1\end{pmatrix}=-4$ //you call it a fish? He can call it a dinosaur if he wanted to :D
solving gives us:
$k=1,l=-3$
so $x=u+1 \quad y=v-3$
$(-3u+v)du+(u+v)dv=0$ //Beautiful! It's homogenous now
$\frac{ dv }{ du }=\frac{{3u-v}}{u+v}$
divide top and bottom by $u$ so we turn the homogenous equation into the form #de_h_type1 and solve it using the tools we developed from lecture 2.
$\frac{ dv }{ du }=\frac{{3-\frac{v}{u}}}{1+\frac{v}{u}}$

$\frac{v}{u}=w \quad v=uw \quad \frac{ dv }{ du }=w+u\frac{ dw }{ du }$
$w+u\frac{ dw }{ du }=\frac{{3-w}}{1+w}$ If you remember from lecture 2, after these substitutions the equation should now be separable, we just move the w terms to one side and the u terms to the other:
$u\frac{ dw }{ du }=\frac{{3-2w-w^2}}{1+w}$
$-\frac{{w+1}}{w^2+2w-3}dw=\frac{du}{u}$ <- Like that :)
$\int-\frac{{w+1}}{w^2+2w-3}dw=\int\frac{du}{u}$
let $z=w^2+2w-3$
$dz=2(w+1)dw$
$\frac{-1}{2}\int \frac{dz}{z}=\ln\mid u\mid$
$\frac{-1}{2}\ln|z|+C=\ln\mid u\mid$
$\ln\mid z\mid^{1/2}+\ln\mid u\mid=C$
$\ln(\mid z\mid^{1/2}\mid u\mid)=C$
$\mid z\mid^{1/2}u=e^C$
> How did he get rid of the abs()? I'm not sure. But he fixes the problem right after:

$\mid z\mid u^2=e^{2C}$
> Funny enough, after that step above of squaring both sides is done, it's like he never even dropped the abs to begin with. All solutions are reobtained again.

$zu^2=A$
> This step I can understand.

$\left( \left( \frac{v}{u} \right)^2+2\frac{v}{u}-3 \right)u^2=A$
remember $u=x-1 \quad v=y+3$
$$\left( \left( \frac{{y+3}}{x-1} \right)^2+\frac{2(y+3)}{x-1}-3 \right)(x-1)^2=A$$
you can "simplify" it to: $(y+3)^2+2(y+3)(x-1)-3(x-1)^2=A$ But we are done.

---